大疆2018校招筆試題

大疆的題，挺複雜，最起碼描述是的。。。

底下的答案僅供參考，沒有一道是全部AC的。

//兩個小車，走一步能量消耗1，方向為1向右，-1為向左，首先輸入路途長度，然後輸入兩行，每行第一個為小車的能量，第二個位小車起始位置，第三個為方向。求幾個小車可以走出去？

#include <iostream>

#include <vector>

using namespace std;

int Many(vector<int> &ve,int len)

{

if(ve.size() < 6 || len <= 0)

return 0;

int end = 0;

int power_a = ve[0];

int local_a = ve[1];

int direc_a = ve[2];

int power_b = ve[3];

int local_b = ve[4];

int direc_b = ve[5];

while(power_a > 0 || power_b > 0)

{

if(direc_a == direc_b)

{

if(direc_a == 1)

{

if(local_a + 1 == local_b && power_b == 0)

{

if(power_a > 1)

{

local_a++;

local_b++;

power_a -= 2;

}

}

if(power_a > 0)

{

local_a++;

power_a--;

}

if(power_b > 0)

{

local_b++;

power_b--;

}

}

else

{

if(local_a + 1 == local_b && power_a == 0)

{

if(power_b > 1)

{

local_a--;

local_b--;

power_a -= 2;

}

}

if(power_a > 0)

{

local_a--;

power_a--;

}

if(power_b > 0)

{

local_b--;

power_b--;

}

}

}

else

{

if(local_a < 0 || local_a > len-1)

power_a = 0;

if(local_b < 0 || local_b > len-1)

power_b = 0;

if(abs(local_a - local_b) == 2 && direc_a == 1 && direc_b == -1)

{

power_a -= 2;

power_b -= 2;

}

else

{

if(abs(local_a - local_b) == 1)

{

if(direc_a == 1 && direc_b == -1)

{

local_a--;

local_b++;

power_a -= 2;

power_b -= 2;

}

else

{

local_a --;

local_b++;

power_a --;

power_b++;

}

}

else

{

if(direc_a == -1)

{

local_a --;

power_a--;

}

else

{

local_a++;

power_a--;

}

if(direc_b == -1)

{

local_b --;

power_b--;

}

else

{

local_b++;

power_b--;

}

}

}

}

}

if(local_a < 0 || local_a > len-1)

end++;

if(local_b < 0 || local_b > len-1)

end++;

return end;

}

int main()

{

vector<int> ve;

int len;

cin>>len;

int tmp;

for(int i = 0; i < 6; ++i)

{

cin>>tmp;

ve.push_back(tmp);

}

cout<<Many(ve,len)<<endl;

}

//一共N種花，插花需要每次選M種，每種R支。第二行輸入每種花個數，求最多有多少種插花方法。

#include <iostream>

#include <vector>

#include <algorithm>

using namespace std;

bool fun(vector<int> &ve,int m)

{

int i = ve.size()-1;

int num = 0;

for(; i >= 0; --i)

{

if(ve[i] != 0)

num++;

else

break;

}

if(num < m)

return false;

return true;

}

int Many(vector<int> &ve,int n,int m,int r,int num)

{

if(ve.size() <= 0 || n <= 0 || m <= 0 || r <= 0)

return 0;

int end = 0;

int i = ve.size()-1;

while(fun(ve,m))

{

int size = i;

sort(ve.begin(),ve.end());

for(int j = 0; j < m; ++j)

{

ve[size] --;

size--;

}

end++;

num -= m;

}

return end;

}

int main()

{

int n,m,r;

cin>>n>>m>>r;

vector<int> ve(n);

int tmp;

int num = 0;

for(int i = 0; i < n; ++i)

{

cin>>tmp;

tmp /= r;

ve.push_back(tmp);

num += tmp;

}

cout<<Many(ve,n,m,r,num)<<endl;

}

//輸入初始位置和結束位置，以及二維數組的大小，與其中的元素，為0可以走，為1，其上下左右不能走，如果為2，則該位置的上兩個，下兩個，左兩個，右兩個不能走。以此類推，求最短路徑？

#include <iostream>

#include <vector>

using namespace std;

vector<int> endpath;

void Get_num(vector<vector<int>> &path,int n,int m)

{

for(int i = 0; i < n; ++i)

{

for(int j = 0; j < m; ++j)

{

if(path[i][j] != -2)

{

int tmp = 0;

for(int k = 0; k < path[i][j]; k++)

{

tmp = i - k - 1;

if(tmp >= 0)

path[tmp][j] = -2;

else

break;

}

for(int k = 0; k < path[i][j]; k++)

{

tmp = i + k + 1;

if(tmp < n)

path[tmp][j] = -2;

else

break;

}

for(int k = 0; k < path[i][j]; k++)

{

tmp = j - k - 1;

if(tmp >= 0)

path[i][tmp] = -2;

else

break;

}

for(int k = 0; k < path[i][j]; k++)

{

tmp = j + k + 1;

if(tmp < m)

path[i][tmp] = -2;

else

break;

}

}

}

}

}

void Less_len(vector<vector<int>> &path,int n,int m,int i,int j,int endi,int endj,int num)

{

path[i][j] = -1;

if(i == endi && j == endj)

endpath.push_back(num);

if(i < n && j < m)

{

if(i - 1 >= 0 && path[i-1][j] == 0)

Less_len(path,n,m,i-1,j,endi,endj,num+1);

else if(i + 1 < n && path[i+1][j] == 0)

Less_len(path,n,m,i+1,j,endi,endj,num+1);

else if(j - 1 >= 0 && path[i][j-1] == 0)

Less_len(path,n,m,i,j-1,endi,endj,num+1);

else if(j + 1 < m && path[i][j+1] == 0)

Less_len(path,n,m,i,j+1,endi,endj,num+1);

}

}

int Less_len(vector<int> &ve,vector<vector<int>> &path,int n,int m)

{

if(ve.size() <= 0 || path.size() <= 0 || n <= 0 || m <= 0)

return -1;

Get_num(path,n,m);

int num = 0;

Less_len(path,n,m,ve[0],ve[1],ve[2],ve[3],num);

int less = 0;

for(int i = 0; i < endpath.size(); ++i)

{

if(i == 0)

less = endpath[i];

if(less > endpath[i])

less = endpath[i];

}

return less;

}

int main()

{

int n,m;

vector<int> ve;

int tmp;

cin>>n>>m;

for(int i = 0; i < 4; ++i)

{

cin>>tmp;

ve.push_back(tmp);

}

vector<vector<int>> path(n);

for(int i = 0; i < n; ++i)

{

path[i].resize(m);

}

for(int i = 0; i < n; ++i)

{

for(int j = 0; j < m; ++j)

{

cin>>tmp;

path[i][j] = tmp;

}

}

cout<<Less_len(ve,path,n,m)<<endl;

return 0;

}

//H的水桶，注水速度X，h處有洞，流水速度Y，S秒以後水深（四舍五入）。

#include <iostream>

using namespace std;

int Time(int H,double x,double y,double h,int s)

{

if(H < 0 || H > 5000 || x < 0 || x > 5000 || h < 0 || h > 5000

|| s < 0 || s > 5000 || h > H)

return -1;

double end = x * s - y * (s - h / x);

if(end >= H)

return H;

int tmp = end * 10;

tmp %= 10;

if(tmp > 4)

return end+1;

return end;

}

int main()

{

int H,s;

double h,x,y;

cin>>H>>x>>y>>h>>s;

cout<<Time(H,x,y,h,s)<<endl;

return 0;

}